2025 Cambridge IGCSE Science - Combined (0653) 模擬試題連答案詳解

(a) Acceleration is calculated using the formula:
\(a = \frac{v - u}{t}\)
Substituting the values:
\(a = \frac{15 - 0}{6.0} = 2.5\text{ m/s}^2\)

(b) Using Newton's second law:
\(F = m \times a\)
\(F = 800\text{ kg} \times 2.5\text{ m/s}^2 = 2000\text{ N}\)

(c) Kinetic energy is given by:
\(\text{KE} = \frac{1}{2}mv^2\)
\(\text{KE} = 0.5 \times 800 \times 15^2 = 400 \times 225 = 90000\text{ J}\) (or \(90\text{ kJ}\))

(d) The electric motor transfers electrical energy into useful kinetic energy.

(e) Work done is the product of the force and the distance moved in the direction of the force: \(W = F \times d\).

(a) [2 marks total]
- 1 mark for correct formula \(a = \frac{v-u}{t}\) or substitution.
- 1 mark for correct calculation with units: \(2.5\text{ m/s}^2\).

(b) [2 marks total]
- 1 mark for correct formula \(F = ma\) or substitution.
- 1 mark for correct answer: \(2000\text{ N}\).

(c) [2 marks total]
- 1 mark for correct formula \(\text{KE} = \frac{1}{2}mv^2\) or substitution.
- 1 mark for correct answer: \(90000\text{ J}\) or \(90\text{ kJ}\).

(d) [1.88 marks total]
- 1 mark for identifying 'electrical energy' as the input.
- 0.88 marks for identifying 'kinetic energy' as the useful output.

(e) [1 mark total]
- 1 mark for defining work done as force multiplied by distance moved in the direction of the force.

(a) Hydrocarbon X is an alkane with 4 carbon atoms, which is butane (\(\text{C}_4\text{H}_{10}\)). The displayed formula shows a linear chain of 4 carbon atoms connected by single bonds, with each carbon surrounded by hydrogen atoms to ensure four bonds per carbon.

(b) To distinguish alkanes from alkenes, add bromine water (aqueous bromine) to both substances in the absence of UV light. Being saturated, butane does not react and the solution remains orange-brown. Butene, being unsaturated, undergoes an addition reaction across the double bond, decolorizing the bromine water from orange-brown to colorless.

(c) Catalytic cracking requires high temperatures (typically \(500^\circ\text{C}\) to \(700^\circ\text{C}\)) and a catalyst such as alumina (\(\text{Al}_2\text{O}_3\)) or silica (\(\text{SiO}_2\)) or zeolite.

(a) [2.88 marks total]
- 1 mark for identifying the name 'butane'.
- 1.88 marks for drawing the correct fully displayed structure showing all atoms (4 C, 10 H) and all single bonds (C-C and C-H).

(b) [3 marks total]
- 1 mark for stating the reagent as bromine water (or aqueous bromine).
- 1 mark for stating that butane remains orange/brown (no reaction).
- 1 mark for stating that butene turns the solution colorless (decolorizes).

(c) [3 marks total]
- 1 mark for identifying a suitable catalyst: alumina / silica / zeolite.
- 2 marks for stating a correct temperature range (e.g., \(500^\circ\text{C}\) to \(700^\circ\text{C}\); accept any value within \(450\) to \(800^\circ\text{C}\)).

(a) Sperm cells and egg cells have distinct adaptations:
1. Sperm cells have a flagellum (tail) for swimming, whereas egg cells are stationary.
2. Sperm cells are very small, whereas egg cells are much larger.
3. Sperm cells contain an acrosome in their head (containing digestive enzymes to penetrate the egg), whereas egg cells have a jelly layer and a large cytoplasm rich in nutrients and energy stores.

(b) Fertilization (the fusion of the nuclei of the male and female gametes) occurs in the oviduct (fallopian tube).

(c) The placenta is adapted for diffusion because:
1. It features finger-like projections (chorionic villi) which greatly increase the surface area available for exchange.
2. The membrane separating the maternal and fetal blood is very thin (only a few cells thick), minimizing the diffusion distance.
3. It has a rich blood supply on both the maternal and fetal sides, keeping blood moving to maintain steep concentration gradients for gases (oxygen and carbon dioxide) and nutrients.

(a) [3.88 marks total]
- 1 mark for comparing mobility/tail (sperm has a flagellum, egg does not).
- 1 mark for comparing size (sperm is small, egg is much larger).
- 1 mark for comparing cytoplasm/nutrients (egg has a large cytoplasm with energy stores, sperm has very little).
- 0.88 marks for identifying specific structures like the acrosome (sperm) or jelly coat (egg).

(b) [1 mark total]
- 1 mark for 'oviduct' or 'fallopian tube'. (Reject ovary/uterus).

(c) [4 marks total]
- 1 mark for mentioning 'large surface area' due to villi.
- 1 mark for mentioning 'thin membrane' or 'short diffusion distance'.
- 1 mark for mentioning 'rich blood supply' or 'continuous blood flow'.
- 1 mark for explaining that these adaptations maintain concentration gradients or increase the rate of diffusion.

(a) (i) For resistors connected in parallel, the reciprocal of the total equivalent resistance is the sum of the reciprocals of the individual resistances:
\(\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}\)
\(\frac{1}{R_{\text{total}}} = \frac{1}{6.0\ \Omega} + \frac{1}{3.0\ \Omega} = \frac{1}{6.0} + \frac{2}{6.0} = \frac{3}{6.0} = \frac{1}{2.0}\)
Inverting both sides gives:
\(R_{\text{total}} = 2.0\ \Omega\).

(ii) Using Ohm's Law for the whole circuit:
\(I = \frac{V}{R_{\text{total}}}\)
\(I = \frac{12.0\text{ V}}{2.0\ \Omega} = 6.0\text{ A}\).

(b) In parallel circuits:
1. Each branch forms an independent circuit. If one lamp breaks or is switched off, the current can still flow through the other branches, so the remaining lamps stay lit.
2. The potential difference across each branch is equal to the supply voltage. This means each lamp operates at its full rated power and brightness, unlike in a series circuit where voltage is shared and lamps would be much dimmer.

(a) (i) [3 marks total]
- 1 mark for correct formula: \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\).
- 1 mark for correct substitution: \(\frac{1}{6} + \frac{1}{3} = \frac{3}{6}\).
- 1 mark for the correct final answer \(2.0\ \Omega\).

(ii) [2.88 marks total]
- 1 mark for the formula: \(I = \frac{V}{R}\).
- 1 mark for substituting \(12\text{ V}\) and their calculated resistance from (i).
- 0.88 marks for correct calculation of current: \(6.0\text{ A}\) (accept error carried forward from (a)(i)).

(b) [3 marks total]
- 1 mark for stating that if one lamp breaks, the others remain on.
- 1 mark for explaining that this is because each lamp is on its own independent branch.
- 1 mark for stating that each lamp receives the full supply voltage (or can be controlled by its own switch).

(a) (i) An exothermic reaction is a process that transfers thermal energy to the surroundings, causing the temperature of the surroundings to increase.
(ii) The energy level diagram should show:
- A horizontal line on the left for the reactants, \(\text{CH}_4 + 2\text{O}_2\), at a higher position.
- A horizontal line on the right for the products, \(\text{CO}_2 + 2\text{H}_2\text{O}\), at a lower position.
- A curve rising to a peak representing the activation energy barrier.
- An arrow pointing up from the reactant level to the peak, labelled \(E_a\) (activation energy).
- An arrow pointing down from the reactant level to the product level, labelled \(\Delta H\) (enthalpy change, which is negative).

(b) During a chemical reaction:
1. Bond breaking is endothermic: energy is absorbed to break the \(\text{C}-\text{H}\) bonds in methane and the \(\text{O}=\text{O}\) bonds in oxygen.
2. Bond making is exothermic: energy is released when the new \(\text{C}=\text{O}\) bonds in carbon dioxide and \(\text{O}-\text{H}\) bonds in water are formed.
Because the energy released when making these new bonds is greater than the energy required to break the original bonds, the overall process releases energy and is exothermic.

(a) (i) [1.88 marks total]
- 1 mark for stating that it releases thermal energy / heat.
- 0.88 marks for stating that this energy is transferred to the surroundings.

(ii) [4 marks total]
- 1 mark for drawing reactants at a higher energy level than products.
- 1 mark for a curve rising to a peak and falling to the products.
- 1 mark for correctly labelling reactants and products.
- 1 mark for correctly labelling \(E_a\) (pointing up from reactants to peak) and \(\Delta H\) (pointing down from reactants to products).

(b) [3 marks total]
- 1 mark for stating that bond breaking requires energy.
- 1 mark for stating that bond making releases energy.
- 1 mark for concluding that more energy is released in making bonds than is absorbed in breaking bonds.

(a) The balanced chemical equation for photosynthesis is:
\(6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light / chlorophyll}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)

(b) (i) The independent variable is the factor changed by the investigator (the distance of the lamp, which alters light intensity). The dependent variable is the factor measured (the rate of bubble production / number of bubbles per minute).

(ii) At low light intensity, light is the limiting factor; as light intensity increases, the rate of photosynthesis increases because more energy is absorbed by chlorophyll. However, at very high light intensities, the rate levels off. This is because light intensity is no longer the limiting factor. The rate is now restricted by some other limiting factor, such as carbon dioxide concentration or temperature, which is in short supply.

(a) [2.88 marks total]
- 1 mark for correct chemical formulas of reactants (\(\text{CO}_2\) and \(\text{H}_2\text{O}\)) and products (\(\text{C}_6\text{H}_{12}\text{O}_6\) and \(\text{O}_2\)).
- 1.88 marks for correct balancing coefficients (6, 6, 1, 6).

(b) (i) [2 marks total]
- 1 mark for identifying the independent variable as light intensity (or lamp distance).
- 1 mark for identifying the dependent variable as the rate of bubble production (or number of bubbles per minute).

(ii) [4 marks total]
- 1 mark for stating that at lower light intensities, light is the limiting factor.
- 1 mark for stating that rate increases with light intensity because chlorophyll absorbs more light energy.
- 1 mark for stating that at the plateau, light is no longer the limiting factor.
- 1 mark for identifying another factor that has become limiting (e.g., carbon dioxide concentration or temperature).

(a) (i) The balanced equation is:
\(\text{Fe}_2\text{O}_3\text{(s)} + 3\text{CO(g)} \rightarrow 2\text{Fe(l)} + 3\text{CO}_2\text{(g)}\)

(ii) The substance reduced is iron(III) oxide (\(\text{Fe}_2\text{O}_3\)). In terms of oxygen transfer, reduction is defined as the loss of oxygen. Here, iron(III) oxide loses oxygen atoms to form elemental iron.

(b) Limestone (\(\text{CaCO}_3\)) undergoes thermal decomposition in the hot blast furnace to produce calcium oxide (\(\text{CaO}\)) and carbon dioxide gas:
\(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\)
Calcium oxide is a basic oxide. It reacts with silicon dioxide (\(\text{SiO}_2\)), which is an acidic impurity, to form molten slag, calcium silicate (\(\text{CaSiO}_3\)):
\(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\)
This molten slag floats on top of the molten iron and is tapped off separately.

(a) (i) [2.88 marks total]
- 1 mark for correct reactants and products.
- 1.88 marks for correct balancing coefficients (1, 3, 2, 3).

(ii) [2 marks total]
- 1 mark for identifying iron(III) oxide (\(\text{Fe}_2\text{O}_3\)).
- 1 mark for explaining that it loses oxygen.

(b) [4 marks total]
- 1 mark for stating that calcium carbonate decomposes to form calcium oxide.
- 1 mark for the correct decomposition equation: \(\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2\).
- 1 mark for explaining that basic calcium oxide reacts with acidic silicon dioxide.
- 1 mark for the correct slag-formation equation: \(\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3\).

(a) Conduction in a metal like copper occurs via two mechanisms:
1. Lattice vibration: The copper ions at the heated end vibrate more rapidly. These ions collide with neighboring ions, transferring kinetic energy along the lattice structure.
2. Delocalized electrons: Copper contains free, mobile electrons. When heated, these electrons gain kinetic energy and diffuse quickly through the solid. They collide with cooler ions in distant parts of the metal, transferring thermal energy much faster than lattice vibrations alone.

(b) When the hot copper block is placed in the water, the nearby water absorbs thermal energy, causing the water molecules to move faster and spread apart. This causes the heated water to expand and its density to decrease. This warm, less dense water rises. Cooler, denser water from the top of the beaker sinks to take its place. This continuous cycle forms a convection current.

(c) Two key differences between evaporation and boiling are:
- Evaporation can happen at any temperature, while boiling only occurs at a specific temperature (the boiling point).
- Evaporation occurs only at the surface of the liquid, whereas boiling happens throughout the entire volume of the liquid (producing bubbles of vapor within the liquid).

(a) [3.88 marks total]
- 1 mark for stating that particles (ions) vibrate and pass energy to neighboring ions.
- 1 mark for identifying the role of delocalized / free / mobile electrons.
- 1 mark for explaining that these electrons gain kinetic energy and move through the metal.
- 0.88 marks for explaining that electrons transfer energy by colliding with other ions/electrons.

(b) [3 marks total]
- 1 mark for stating that heated water expands and its density decreases.
- 1 mark for stating that the less dense, warm water rises.
- 1 mark for stating that cooler, denser water sinks to replace it, creating a circulating current.

(c) [2 marks total]
- 1 mark for contrasting temperatures (evaporation at any temperature, boiling only at boiling point).
- 1 mark for contrasting locations (evaporation only at surface, boiling throughout the liquid/forming bubbles).

(a) Using Newton's second law: \(F = ma\). Rearranging for acceleration: \(a = \frac{F}{m} = \frac{15\text{ N}}{2.5\text{ kg}} = 6.0\text{ m/s}^2\).

(b)(i) Using the equation of motion: \(v = u + at\). Since the trolley starts from rest, \(u = 0\). Therefore, \(v = 0 + (6.0\text{ m/s}^2 \times 4.0\text{ s}) = 24\text{ m/s}\).

(b)(ii) The formula for kinetic energy is \(E_k = \frac{1}{2}mv^2\). Substituting the values: \(E_k = 0.5 \times 2.5\text{ kg} \times (24\text{ m/s})^2 = 1.25 \times 576 = 720\text{ J}\).

(c) The kinetic energy of the trolley is entirely converted into heat energy by work done against the average decelerating force. Work Done \(= \text{Force} \times \text{distance}\). Here, Work Done \(= \Delta E_k = 720\text{ J}\). Therefore, \(720 = F \times 8.0\text{ m}\), which gives \(F = \frac{720}{8.0} = 90\text{ N}\). Alternatively, using equations of motion: \(v^2 = u^2 + 2as \implies 0 = 24^2 + 2 \times a \times 8.0 \implies a = -36\text{ m/s}^2\). Then, \(F = ma = 2.5\text{ kg} \times 36\text{ m/s}^2 = 90\text{ N}\).

(a) [2 marks]
- \(a = \frac{F}{m}\) or \(15 / 2.5\) (1 mark)
- \(6.0\text{ m/s}^2\) (1 mark)

(b)(i) [2 marks]
- \(v = u + at\) or \(a \times t\) (1 mark)
- Correct substitution shown resulting in \(24\text{ m/s}\) (1 mark)

(b)(ii) [2 marks]
- \(E_k = \frac{1}{2}mv^2\) or \(0.5 \times 2.5 \times 24^2\) (1 mark)
- \(720\text{ J}\) (1 mark)

(c) [3 marks]
- Work Done \(=\) Loss of \(E_k\) or \(W = F \times d\) (or suitable equations of motion to find deceleration, e.g., \(v^2 = u^2 + 2as\)) (1 mark)
- \(F = \frac{720}{8.0}\) (or \(F = 2.5 \times 36\)) (1 mark)
- \(90\text{ N}\) (1 mark)